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I Help with Euler-Lagrange Equation

  1. Jun 25, 2016 #1
    I have begun teaching myself Lagrangian field theory in preparation for taking the plunge into quantum field theory ( it's just a hobby, not any kind of formal course ). When working through exercises, I have run across the following issue which I don't quite understand. I am being given a Lagrangian density, and ask to derive the equations of motion; I understand the principles involved, and everything is fine and easy until I get to the point where I need to evaluate the following expression :

    [tex]\frac{1}{2}\frac{\partial }{\partial \left ( \partial _{\mu}\varphi \right )}\left ( \partial _{\mu}\varphi \right )^2[/tex]

    wherein ##\varphi(x,y,z,t)## is a scalar field. My approach to this was as simple as it was naive :

    [tex]\frac{1}{2}\frac{\partial }{\partial \left ( \partial _{\mu}\varphi \right )}\left ( \partial _{\mu}\varphi \right )^2=\frac{1}{2}\frac{\partial }{\partial \left ( \partial _{\mu}\varphi \right )}\left ( \partial _{\mu}\varphi \right )\left ( \partial _{\mu}\varphi \right )[/tex]

    which evaluates to ##\partial _{\mu}\varphi## via the product rule. However, this is where I get stuck, because the answer is wrong - the correct approach should have been

    [tex]\frac{1}{2}\frac{\partial }{\partial \left ( \partial _{\mu}\varphi \right )}\left ( \partial _{\mu}\varphi \right )\left ( \partial ^{\mu}\varphi \right )[/tex]

    which apparently evaluates to ##\partial ^{\mu}\varphi## ( though I have difficulties with that as well, but that's a separate issue ), and leads to the correct equations of motion. My question is : why is ##\left ( \partial _{\mu}\varphi \right )^2=\left ( \partial _{\mu}\varphi \right )\left ( \partial ^{\mu}\varphi \right )## and not ##\left ( \partial _{\mu}\varphi \right )^2=\left ( \partial _{\mu}\varphi \right )\left ( \partial _{\mu}\varphi \right )## ? I know that this is probably something very elementary, so please don't laugh at me, but I genuinely don't get it.
     
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  3. Jun 26, 2016 #2

    stevendaryl

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    This is not specific to quantum field theory, but is a fact about vectors and covectors.

    A vector is something that is generally written with upper-indices. For example, if a massive particle's position is [itex]x^\mu[/itex], then its 4-velocity is given by [itex]\frac{dx^\mu}{d\tau}[/itex] (where [itex]\tau[/itex] is proper time for the particle), which is a vector. In contrast, a covector is written using lower-indices. For example, if [itex]\phi[/itex] is a scalar field, then its "gradient" is [itex]\partial_\mu \phi \equiv \frac{\partial \phi}{\partial x^\mu}[/itex].

    To make a scalar, you have to use a vector together with a covector. That means that you have to have one object with upper-indices and another object with lower-indices. So if [itex]A^\mu[/itex] is a vector and [itex]B_\mu[/itex] is a covector, then you can make a scalar by combining them via: [itex]\sum_\mu A^\mu B_\mu[/itex]. (It's usually written without the [itex]\sum[/itex], using the convention that repeated indices, one upper and one lower, are always summed over.) For example, we can combine a 4-velocity [itex]\frac{dx^\mu}{d\tau}[/itex] and a covector [itex]\partial_\mu \phi[/itex] to get the combination [itex]\frac{dx^\mu}{d\tau} \partial_\mu \phi[/itex]. This is a scalar, the rate of change of [itex]\phi[/itex] for the particle as a function of proper time.

    So if you have to use a vector and a covector to make a scalar, then how do you take the square of a vector? The answer is that you can't take the square of a vector without help from a metric tensor. The metric tensor [itex]g_{\mu \nu}[/itex] is an operator that converts a vector into a covector: If you have a vector [itex]A^\mu[/itex] then you can convert it into a corresponding covector [itex]A_\mu[itex] through [itex]A_\mu = g_{\mu \nu} A^\nu[/itex] (by convention, the repeated index [itex]\nu[/itex] on the right side of = is summed over). Using the metric, we can square a vector as follows:

    [itex]|A^\mu|^2 \equiv A^\mu A_\mu \equiv g_{\mu \nu} A^\mu A^\nu[/itex] (in the last expression, both [itex]\mu[/itex] and [itex]\nu[/itex] are summed over).

    Similarly, the square of a covector must involve a metric, as well:
    [itex]|\partial_\mu \phi|^2 \equiv \partial^\mu \phi \partial_\mu \phi \equiv g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi[/itex]
    (where [itex]g^{\mu \nu}[/itex] is the inverse of [itex]g_{\mu \nu}[/itex]; it converts a covector [itex]\partial_\mu \phi[/itex] into a vector, [itex]\partial^\mu \phi[/itex]).

    In quantum field theory, if an expression like [itex](\partial_\mu \phi)^2[/itex] appears in the Lagrangian, it always means [itex]\partial^\mu \phi \partial_\mu \phi[/itex], because you can't take the square of a covector, otherwise.
     
  4. Jun 26, 2016 #3

    stevendaryl

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    Oh, one further point. Since [itex]\partial^\mu \phi \partial_\mu \phi[/itex] means [itex]g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi[/itex], there are two factors of [itex]\partial_\mu \phi[/itex], so taking the derivative with respect to [itex]\partial_\mu \phi[/itex] gives a factor of two.
     
  5. Jun 26, 2016 #4
    My goodness, of course !! I had come across that already in my studies of GR, but it had completely slipped my mind in this context. I knew I was missing something elementary. Thank you for taking the time to reply in such detail, I understand it now :woot:
     
  6. Jun 26, 2016 #5
    Yup indeed :) Given the correct expansion of this bracket, I actually got the rest of the exercise done with no problems, it was really just this one upper index that tripped me up. But all is good now :smile:
     
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